Question

Show that |a-b| greater or equal to ||a|-|b|| for all cases a, b in real numbers

Answer 1

I3+8I=5 and II3I+I8II=5

Answer 2

so it is the same

Answer 3

in other words equal

Answer 4

Using the property: \(|p| = \sqrt{p^2}\)\[\ \ \ \ \ \Big||a| - |b|\Big|\]\[= \sqrt{\left(|a| - |b|\right)^2}\]\[ = \sqrt{|a|^2 -2|ab| + |b|^2}\]\[=\sqrt{a^2 - 2|ab| + b^2}\]
\[\ \ \ \ |a-b| \]\[= \sqrt{(a-b)^2}\]\[=\sqrt{a^2 - 2ab + b^2}\]

\[\sqrt{a^2 - 2|ab| + b^2} \le \sqrt{a^2 - 2ab + b^2}\]
\[\implies \Big||a| - |b|\Big| \le |a-b|, \forall (a,b) \in \mathbb{R}^2\]

Answer 5

In case it wasn't clear, the reason is that \(-2|ab| < -2ab\) if only one of a or b is negative. But if they're both negative or both positive, then \(-2|ab| = -2ab\)